(C) Since the impulsive force acts at the pivot,the angular momentum about the pivot is conserved during the collision.Initial angular momentum $L_i =

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(C) Since the impulsive force acts at the pivot,the angular momentum about the pivot is conserved during the collision.Initial angular momentum $L_i = m v L$.Final angular momentum $L_f = I_{total} \omega$,where $I_{total} = I_{rod} + I_{particle} = \frac{M L^2}{3} + m L^2$.Equating $L_i = L_f$:$m v L = \left( \frac{M L^2}{3} + m L^2 \right) \omega$Substituting the given values $(M = 0.9\, kg, m = 0.1\, kg, L = 1\, m, v = 80\, m/s)$:$0.1 \times 80 \times 1 = \left( \frac{0.9 \times 1^2}{3} + 0.1 \times 1^2 \right) \omega$$8 = (0.3 + 0.1) \omega$$8 = 0.4 \omega$$\omega = \frac{8}{0.4} = 20\, rad/s$.

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

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$A$ thin rod of mass $0.9\, kg$ and length $1\, m$ is suspended,at rest,from one end so that it can freely oscillate in the vertical plane. $A$ particle of mass $0.1\, kg$ moving in a straight line with velocity $80\, m/s$ hits the rod at its bottom-most point and sticks to it. The angular speed (in $rad/s$) of the rod immediately after the collision will be

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

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